# Solve for x lg(x+1)-lg9=1-lgx

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We'll re-write the equation, moving the terms in x to the left side and the terms without x, to the right side.

lg(x+1) - lg9 = 1 - lgx

lg(x+1) + lgx = 1 + lg9

We'll re-write 1 as lg 10

lg(x+1) + lgx = lg 10 + lg9

Since the bases are matching, we'll use the product rule of logarithms both sides:

lg x(x+1) = lg 90

Since the bases are matching, we'll use one to one rule:

x(x+1) = 90

We'll remove the brackets:

x^2 + x - 90 = 0

We'll apply the quadratic formula:

x1 = [-1+sqrt(1+360)]/2

x1 = (-1+19)/2

x1 = 9

x2 = (-1-19)/2

x2 = -10

**We'll reject the second solution, because it's negative. We'll keep the solution x = 9.**

To solve for x: lg(x+1)-lg9=1-lgx

log(x+1) - log9 = 1-logx

log(x+1)/9 =log10 - log(x) , as loga=logb = loga/b.

log(x+1)/9 = log10/x.

We take antilog:

(x+1)/9 = 10/x.

(x+1)x = 10x.

x^2+x= 10x.

x^2-x-10x = 0

x^2-9x = 0.

x(x-9) = 0

x = 0 or x-9 = 0.

x= 0, or x=9.